# Necessary conditions for “convergence in mean” of a sequence of integrable functions.

Let $(X,\mathscr F,\mu)$ be a measure space (not necessarily finite) and $\{f_n\}_{n\in\mathbb N}$ and $f$ be nonnegative, real-valued and integrable functions on $X$ such that $f_n\to f$ almost everywhere. I'm looking for necessary conditions to have "convergence in mean", i.e., $\int |f_n-f|~d\mu\to 0$. First, two remarks:

(1): If there exists an integrable function $g:X\to[0,\infty]$ such that $f\leq g$ almost everywhere and $f_n\leq g$ almost everywhere for every $n\in\mathbb N$, then Lebesgue's dominated convergence theorem implies that $\int |f_n-f|~d\mu$ converges to zero, since $|f_n-f|\to 0$ almost everywhere and $|f_n-f|\leq f_n+f$ is dominated by $2g$ almost everywhere.

(2): If the $f_n$ are not dominated by an integrable function, then the convergence in mean can potentially fail: Take $f=0$ and $f_n=\chi_{[n-1,n]}$ for instance ($\chi$ denotes the indicator function).

My question is the following: If, instead of assuming there exists an integrable function $g:X\to[0,\infty]$ which dominates $f$ and the $f_n$ almost everywhere, we assume that $\int f_n~d\mu$ converges to $\int fd\mu$, then does this imply that $\int |f_n-f|~d\mu$ converges to $0$? So far I have not been able to prove the result directly, or find a counterexample: Every counterexample I can imagine is of the "form" of my remark (2), that is, such that $\int f_n~d\mu$ does not converge to $\int f~d\mu$.

Any hint/useful remark/counterexample would be greatly appreciated.

Indeed, given that $f_n\to f$ almost everywhere, a necessary and sufficient condition for $\int_X|f_n-f|\mathrm d\mu\to 0$ is $\int_X|f_n|\mathrm d\mu\to\int_X|f|\mathrm d\mu$. See here or there for a proof.